metrology

1. The d -,sofatr has been in connec
1
II
f ,
<
Complete interchangeability in the
ection and materials hqdling by usi r
ded according according tb ,i .method is especially useful i ',i
! Its in complctte prottxtion C 0
!
I
!
1,1 .
f a axeztain-tqrgeafpiston
skirt each be O M mm.
3;". '
tnr = 0.04)ildm P*
clearance = largest bore-smallest piston
k g by selective
bsrltsmdomxraeethat
I, of tb d e r of 2.5 thou
more I-- slnd'gradd W
s. Only the inside and
ens. An incidental h n t a g e of this
suitable for various purposes b
different gnorrfrs. W+at means, three desirable classes are dri
bewing, a snug b@ lfilee bearing, a d a preloeded &in@ for
lspidb.
9.7. WLVED EXAMPLES :
. .
. .. G.mpb 1. ~ j n dthe values of allowance, hole tolerancd and s w t tolerance fw t&
i . bollwing .&hhm.yjoniqf mated parts -according,do basic h ~ i eqvNe~.
4P,,r' H Q ~: 47.5Qmm Shaft :37.47 mm, I .
37.52 mm 37.45' mm . , f
'.I j

2. = 37.52,- 3130 w8 s a d s& hi+,trl;s13 >r
= 0 . 9 rmR.= &&3Rti,@ y&$&J
Shaft tolerrtrrce = High Jilrait - m?WritbM,
--
= 3?*4,7 -37.45
' O . ~ ~ = ~ ~ ~ri;;*:
Allowance.=,Max,@- meQal & d m of hale
Example 2. A 75 mm,&@, r
is 0.075 rrun and the required ailowan
the bear&$ bore with the basic,b&.
Solution. Refer to Fig. 9.15
It is clear with the'kit'
E u a p k 3. A medium force f& on a 75

3. It is clear with the basic hobi& thai-; " i8
. ,. . LOW limit of hole = B‘%w "
~ i & =gy&&~.& &ad;, <, ?+I @H+
'' . -e?-. ls;.2&8 inm ~ , a l ,
High lhit of &a = 732& '+ 0.225
sm xu -3s : b ~ f ~w . m wwdik (ikik) *
' ~xupl..&&&@$&-n'-ert?~i~&an and tolerances and hence the limits ofsize
for the shafifW W# +$&! ~in@~~W&f/.The diameter steps are 50 mm and
80 mm. TD.T{ t!?
(b)We know that for hob;&& &v-7 ismfIT3w%,&ndamentaldeviation
Shaft :Hi& limit = -sincetheshaft 'j'liesbelow the mro
*tab m 9ld qwq>q

4. = 59.37 -0.03 = 59.94 mm
The fit is shown in Fig. 9.17. Jt. is a clearance fit with &Or93 @;&J&~..~QIqwance.
Example 5. In a limit system the jhllowirtg limi& we sgwfled to give a clearance fit
between a shaji and a hde :
t - ? dod fo firnil @if+ ,
Determine :(a) ~asi=s&e (b) sh@ and hde to era&& (3the shajl and hole limits (4the
maximum and minimum clearance. "
, I : , ? ..[. c y : , ! I@$:>$ .$ , , -:
.7
+ . + : k t ,;.a,8 -$<$ -,-I , *!L.
Solution.
(a)
. 4 l - l ' ~ - ~ T l # ' I ,,
~ k i c ' s i z e= 30 mmr
(b) SWtolerance = 0.018 1'0.0b~jGb.013 mm -.<
m .~l-,ij If. i:!: r',: t ,
Hole 'tolerand 0.020 km
.'.acir $im* J x & rtk '- I i t ! . 1 % i t ~ , ,L
~ i g hr i i t of shd = 30 - o,@$ r ! ~ " * * .-4
-,',,. 7 , : . t , . -,>sf$,,'>3 -
= 29.995 m%?::r:c 5 . I t :,, . *,!,*. ., ~maul*
Low limit of shaft:, 30 - 0.018
-29.982 ,mtq ,
. ' P ! ' 7 8 . : , { * I .,
High limiti'bfhole ='30 + 0.620-
5(,@20~,iitm
Low limit of hole = 30 mm
< ,I .
Maximum clearance = ~ i ~ hjimit .di.h& - low limit dfshad
~ i n ' i m scl& = Low limit of hole - High limit of shah
Examp@&A hble k d shaft &e a basic site of 25 mm, a)td are to have a clearance$it
with maximumlolearaneeof.O.02 mn a d a mi~limumclearance of 0.01 mm. The hole tolerance
is to be 1.5 timdp thesk@ &lev=+. &itermine :limitsfor both hole and shafi (a)using a hole
bast3 system (6) using a shaji bash wgtp- .
Solution. ~ e f a gto Fig. 9.18.
I . k l :'7 ::,2 nc.r.;~s$mrCt
Fig. J.P8 .,
1'. ,6
If x is the shaft Wmuice Imd y is the hole tolerance, then

5. A Textbook of ProductionEngirieeting
.. x = 0.004 mm and y = 0.006 mm
(d)wok basis system. The lower deviation is z;ero, 71.9 2t;i I!! r r ~ ~ d fy i l i t
*
3 t 1 > Low limit of hole ~t25 mm (Basic size) v,$$?[ "r
.? %($Ziti? >f
%7*Ju !+fin ibA-8. k (,I.- : I .
High limit of hole = 25 + tolerance
= 25.006 mm ZWD--.
m a , , , Qt I%'?,
Upper (High) limit of shaft = low limit of hole - minimum clearance
Low limit of shaft = 24.99 - 0.004 = 24.986 m6-- '-
-": ,, , . . ~ . G I .x's :* t 9 , :lb0k&. *
(b)Shaft basis system. The upper deviation is zero. A3wmi. .-. High limit of shaft = Basic size = 25 mm ,.oiwi&
Low limit of shaft = 25 - 0.904 yi3439fpm
in)
L y limit gf hole = 25 + 0.01 =,22i0a,mmGnd ;,<.)
~i~h"limP'ofhole = 25.0 1 + 0.006 = 25.016 ,mm
Example 7. A jil is designafed as 100 G,/ q,Find the diYqg1h2&~he hole and the shafl.
The diameter steps are :80 rnm and 120 mm.
Sohtioa. Geometric mean diamete{ 6 lr-li, 30 tim11wOJ= 9 s m M
. .
Tolerance grade, i = 0.45 (D): io . ~ I - D ,microw
' ; I d to dmil 1171'1
-
- 0.45t9qi .@,$@I x 98
'
= 2.079 + 0.098 p 2.177 micronsI t l ( . . 8 1 m
> . .
Now, for hole GRi, tolerance = 16i = 34.832 microns ,L,t,
Itant 1.3 11 w w I I I S )'I: ". '
= 0.035 mm (rationahzed)
9' -- Or,Q.fbC
For shaft 4, tolerancemz&~a= 54.425 microns
- 0.054 mm (rationalized)%&7n titmil d+rU - 4 4 4 b~~ t i f t t iW O , ~
We can get these values of tol~mcesdimfly from Table : Appendix IE.
Now F.D.~~wh& G,;; + 2.5 microns (T 9.2)
&it3 :-l>%a~'$- : US^ ,US>Q fliV r-: +. 5 3 . >!
= 25 (98)OJ4= 0,012 nyn (rqtionalized)
.un,w&ihl,fir shahm Q.a"fg Y ~ ~ ~ a ~ ~,RUIR?U T Id DO". mimm (7 9.2) ,,fib '=15, .;I .
L) h . U i > $7i :-yli.~?)SD 9 h A lffUQ wsim'A .tmm-aq1 (gsp11,m i c t o ~ w ' :XI t . ' m . 2 : 1.. ,.:
Dimension See F 9,!3.
(i) Hole - d--- - - - L.L. of H& &wic-#ze + F.D.
81,. C C ,, = 100 + 0.012 mm = 100.012 mm
.
4 i/mM.L. .of Hole = L.L. + Tolerance
' f
= 10&012 + 0.035 = 100.047 m
(ii) Shaft U.L.or H.L. Sbft r Basic Size - ED.
= 100 -O,Q,72= 94928 mm ,, ,. ,
L.L. of Shaft - H.L. - Tohw;slaFe i
I

6. Example8. A 100mm diameterjournal and bearing assembly has3 clea~.mcefir,with the
iQiqwing- specifcatiotp, ;,,!, , , , , +' I -,.;, , . I * I
Tolerance on bewin$ = 4005 a m t ,,,
sivrl 30 irniii 1awo.f AClew~nce T * 0 2 mm
Determine the si;m sf the bewing ad!he jo~fz#,op~(i)Hole Basis System (ii) Shafi Basis
System. Take Unilateral System of tolerances.
Sobtion. v
,t *. : .
(a) Hole-Basis System : Refer pig. 9.13,
- 2 - f
LOW& limit of sd&r!hg =''&ifs Size - 100 dm'' ' '&"
Highest limit iif ~ k r i n ~ ' ' & ~ L ; ~ .of Bearing +Tekrancd'
"@ 100 +b.005 = 100.005 mm
33n814f~~J I P - ~ : ~:qqU = f i s h 'to nrnr; ~~twc.,Refer to Fig. 9.2,
Higher limit of Joumd k: W&'limit of bearing - allowance
.mar ~14KW0.002
-5 99.99811~11i~tzf2d t d 116d? f r i l
Lower limit afd m a l = Migh.limit -T @ b a h ~~sqql;
43i11;.t410r- .= 99,k)98-0.004;', ?L gqiI X W ~ ~
,,~mid.@@- = 99.W1m
(h) Shgt$+@kJ rSI#s@.w,fiver Fig-9.13
Upper limjt sf.hucnal = BasicSize = 100 mm
Lower limti&<!purna1. = Uppcr, .limit - T~krance! lw+G
= l ~ - p . o o 4
awn 2IQ.W- Z 1: '
-..::,. 7 r ~ i ~ ~ t a&Z &kwer limitof Bearing i.A
Example 9. In an assembly of two matingp- of.IOQ,mmh a ' size, the48 is Inter$erence
and the inter$erence variesfrom 0.05 mm to 0.12 ,mm. .qe tolerance on the two matingparts is
equal. Determine the sizes of the two mating part5 oh (a) Hole Basis Systein (b) Shaft Basis
System. m Wf = x i . = 4 7 : 13 i
Solution. 52rtsrylrfi I tr~r;i[1. :vr,l ' .jti1)4 nl. :q?j
(a) Hble Basis System :Refer Pig. 9.tsF :.<I .: T qdlli
ftdz $m,Lower limit of h o b f i ~M j c siza-=JOO mrmn,,:
Now Refer to Fig. 9.5 (a), -, .q, 6 , ; , : , J i - - I
Maximum interference will be when the hole is at its lower limit and t e shaft.isat its upper
limit,
I
ii

7. :. Upper limit of shaft = Lower limit of hole + Maximum interference
Now it is clear that, t, c . . ,>-. + L~ ' ::, c 7
Maximum interference - Minimum interference= Up-r limit of shaft - Lower limit of
hole - (Lower limit of shaft - Uppet limit of bole3 3 m,
= (Upper limit of $haft-Lower limit of shaft)
+ (Upper limit of hale - Lower limit of hole
;,,t~,d71, , ' ,, :-.,I traitk br?f.rt Tolerance on shaft + Tofirance on hole "{' ""'"' "'"
,,.., ..,,= , ,.. . ... ,, .../I$ 'J1112 :
= 2T .;*1:$2ff$*l=-
2T = 0.12 - 0.05 = 0.07 q9#. ,,,,, .% ,..,.%a-,b)r I
:. Tolerance ~p qbft = Tolerance on hole = 0.933 nun
,.. Hi* l j i o f hole.r , Lower limit + Toleranqt. ,,I I,.,
r,;m ~ ~ J I Iooi = i 0 1 ~100.035 mm
Lower limit of shaft = Upper limit - Tolerance
'-= 100.QSS mm.
(b) Shaft Basis System : i NQCY .
Upper limitdshaft = B d ~ c i z e=.10Ornia;rnr! .:
Lower 'limit of shaft Upper[limit- Tolerance
* 100-0.035 = 99.965 mm
Lower limit of hole - Upper limit of u'-lhxinilrrm ~ " '
awl I'w; = 100-0.12 = w:$8 hutr. : Iyq'J
..-.,1
:. Upper liniit ofhole = Lo& limit'%&kbhke
i ' l s t i r / . ~
* 1 C ' 1
= &.838'+ 0.035 = 99.915 mm.
,ak ,j:je *'?
Example 10. For a number of mterchangeable mathgpms (holes andshqfii), the average
allowance is 0.04 ~ l w nand t k allowance must nd exceed *0.012mm- the merage value.
lRe bark*she 168.rrmr.l@ i e ~ &o#;hdc+ $ ' ~ - t b b & d~ l f A sshap. Lktermine the sizes
of holes and Skofts using Hole h i s -~ 4 e mand Unilateral system of tolerances.
Solution. Refer to Fig. 9.2 L.OO.mt :
Maximum ahitmce --Mhu'mtarn allomot *' l b h w c on shaft + Tolerance on hole
I
,' .n't'i>w.,te. ?'amce &f =t 6.0011 m.4 j5r:,,~': I : N ~i t . '.',? 3u.b v., 7"..p ~hfSns.r.3
, . . .. ,. ':),I .!.: ,.v:<-. ,. ' 5 .~.i$,*.v.yi . Q,.;.-!T i ~ + . t l , ~ . ,3,, 'c,~,,:;
a T o b r a q o n m l e = P.016- , . ~ a i c . . . s .tn ;f.i:-l?,,;; .,;!n,:..,5;~,c.,;.' .'.S.;r I . tr..,
f Lower limit of Hole = Basic Size = 100 mm ':C ..,, . '
Upper limit of Hole = Lower limit + tolerance
Now Min. allawake = Lowc2r%mii ofhde -'Upp&rh i t ofshaft.

8. mrw,mJ~'~&MU ~~~~.!hiitnilB l m @ mrnGGA w ~SIEO~JR* x: kqA
= 99.972 -0.008 lil srij ':c
IWb3 & .,a&!$ Zl Jfltr8 99.964 mmi I Wi fl R* slslfi~01 p i fk;c;i b9ntrfi r+ .(fS
msr .nmrsllm b #KT .ntm (. .,. ,, wtt Lm riod &'*L enoivrvmib '
f { 3 1- $3 -, $-~~IP~QPLEMSs l.nnosarrR m r. bhr
1. Define "Interchangeability" and dimmsiirs IiapbFtancdQ?-
2. Define tolerance. .&I,at ?~[!FIJIL~3ris .T ; J J ~ . , L ~ ~ Iit, 1
l 1 ~ f 3 ga 9% ibl if'~nwxsible€aobtain.c#vexmd-.~w~ ~ l a ~ o lOJ t t .c.:.q rdCi nA .IS
Ea. i s q m r b ) . the w % 3 . wl&'&edt&csbf fit.& 10 ~fl"t&~to~!l:
' ._ I
t b i c l !I
, .r.
5. Define : aflowandi, clearanck ah htederenci: '-;"'
.-..-.....vst: b~a~~mtnd
. 6.' What is zero line? - f # , !; 17, I a,# ?,q-
7. Define : upper i&vh&t,%.&%~r &&&e~#sv31i;f;r;!!4~ 1tm: t "!fl
8. kxpranuun~ad&jj'&;ibitoI* ' ; B i k ~fii'3Ur . .ST
9. What is meant by H&,d&~*pl'&&q4'.~Lqql 3mmm 3m 2-/;3 .LT
10. Explain and compare "Hole basis sy&~'sda'4biKdas
( ' # t r d ~ * ~ g & % ~ ~ ~ ~ ~ ~ . p i ~
iti%iIk- .llker&.dtohave
imum toral cl-. FS #iWW$k~W?&hi(tgtlili?Iautince &a69 Irihrm. Wermine
-
the spccificaticrns of the parts k)i@ @Q& i@..ryld(&).for SratiW average
inmcbngesBil#y.
I S : 611 2 ti ,>L a$;>
12. How will you write the Rt :
is 40 rHm?
.&4rq&$* &basic size
i* 'qg ule &sieuria & * d - . m *I* m amiwii+ arlf 04
0 mm H - h ~ & t h etolkc; grade IR ULO+
P;&!F&,.& t&ga'&ti# ( ant f.O_ 0: 1sl31~1~lb)thdr 4 ilf
($13 the above fig- -. , ., *',
, -: Msiy
t@$qns& e@jpthq.liWofdze for hole
.9: - 7 % *@1&!4$ ~ ~ I P T A
?- tJ.wl D ti1 s~uw@3$&A,,
(b :&A qy r v &;&--;1w,a7.1
e to h c e g d i ~ b25i and f ~ ; ~ n i b e r~'~uaIi&$- I .,, , .itqa-l .+.
15 A gear ring of85 mm diameter bore is died bn baabubiresulting ih'a Ha6 h ~ . l e ?&
tolerances and hence the limits of size for tht h b 'kt& he g& h e .S@$@ttf;t type of fit.
The diamcter stepsare 80 mm and 100mm.'Ihc 8- fWj.&is 0.- mm.
16. Calculate the fbndamental deviations and tdgwn6~1~#amtlW: ,$Qr shaft and
hole p i r - + & n a t t as 60 mm HimCThe tol-ce unit*is g j ~ p la
:I,
. i = 0.45 + 0.00i D mi .scsn&~ 3!{~
The diameter steps are 59 m,@ BB mw!Tka: @I ldg~x@&m, for.m shaft is
= + (IT7 - IT6). For quality 7, the multiplier is 16'and that for quality 6 it is 10.
17. Calculate the tolerances, limits and ~lltowuxcsfor a 25 -4% 'kid hole pair wigmted u
H,ld,. The diameter steps arc 24 mrn snd 30 mm.The multipl(d.kt quality 8 .is25. The FD
for 'd' shaft is - 16PUmicrons Name the type of fi& jh.9 ,,/ &18. Determine the types of fits p~odUCdaby the following maEg &holes anh

9. A Textbook&~ProductionEngineering
19. A fit is designated as :60mm H,2 h,; D b t e n r r ~ hh hi mum'&?&anct! and m9rnum clearance
' "'"- of the tit.
.C
rn0.d s$@:@ ='
lo. A turned shak is to rotate in a reamed h * m a 8 ~ r d fit is HJc,. ~eterhinethe actual
dimensions of the hole and the shaft. The ba#c size is 60 mm. The diameter steps are SO mm
8nd 80 mm. The fundamental deMIdA.w]t&7.sh& c is,
-(95 a)mi )wts :fhlFsl$~y*~ml.* suiifl .t
For quality 7, the muitiplicr is 16.
21. h idler gear is to rotate over a mm s k i , .The:-=fit 'LH#p Dea?rmik(tttk actual
dimensions of the shfdtPIlQ tbbPnt Q$ the i#k.T?w&mew, stcpgm39:and 40 mm. The
hndmental deviation for the hole H is ~ r o*,?tQ3f;?r $9 sh& 5.iq
I 8 :
-(95 + 0.8 D)microns. .,:I 0 <' .$
- The multipli~, f ~quality 7 is 16 a@ that for q q g.is12$.,.
, . r s -
22. Define and compare :Wdoq , w b l y , st&@,~~1lr . > . . . I . * I e*~J1!b
23. Give some common applicatians of Jqive. . '
2 4 what y,*>*n;,*,pf:yy. h a "..&&:,&+
Y*', - 2 , b ~ 6 r . W . d ,B;II$WQ . ~ ~ lmn,
~marmt9m'
-16 *@y "twmdI&l ida#e @& :a~ h & f f . ~ , m-*76 mt'Itg,_~$w;*!:.*
(6)Tha raigtive mor in the dimension is greater in &aft A.
,@lirfas#ti6rt.mtiii
(;,fat&&& &&&&js~" is g,+&
"1) rr.i i ~ fliwl *w .$I
?tmw $+,& ' :'
(d)The rclatbe error in the dimension is greater in s W B. , . - , ,, !,::.(. (Am.:a)
d- -a) Tmsitien FI~J(@lW&m~eSFit (c) clemec fit (&>Noat (GATE 1993) ( h a :e)
7 eL
-y7. h e fit 0" ( %1'&&$l1&ab) ,< .: 2 is &ifid. as H, - s, The type of fit is &:,";;;
(a)Clearance fit !. - & wning fit (yliding fit)
(c) push fit (transition fif) (d)force fit [interference
L - 1 , %(G#TF 1%) A=- :dl
*a~nthq r p k i i o a. , diq-7 > , d 9% ,, , . I ~ b w d w + g ~21
n! *?da) alJowm isaqual t ~ h i l a t s F a l , C w . b .- k'
m8.31~)akvanec is- to UftiMWl Weran&. I -,,bT
brsfi Yc) efiowa~ceiJMebed& eFtd&iwe. d U 3 .&I. .
sion specified by
the tolerance. .a&% 998) [Ans. :el
I 39. A journal md baing aSsOLrrbly has h e il&#bg sizes%' -9h@fmmb *1' PI u m r t q / I s~qqrsr& 4nr , xu- a .@TI - TTij + --0.001
*t.h d ! ' 4 , .- &f ' ro2 -.'L* % n ~z t i d .(kt3 YIO *ltr.jffia:~ .f l
i : I R WIJS~:~4.W t r m ~ r,m
1
+0,001 &iivq@
Bepriflg : SO
4.002 m% **Determine.: T o 1 6 on Journal, Toleran ,Maximumclearance
and type of fit. t W C $5)
.. i t (Ans :0.001 m%&003 m.-0.001 n w W mm,TmstioPL:fif
.. . ?B-

10. 30. In an assembly of journal and bearing, the basis size is 55 mm. The lower deviation, and upper
deviation for bearing is 0 micron and 4 microns respectively. the corresponding values h r
journal are -3 and -7 microns. Detmine the sizes of journal and bearing and the typeof fit.
H.L. = 54.997 mm
(Aas : - Journal : Bearing H-LA= "*Oo4 mm ,clearance fit).
L.L. =54.993 mm ' L.L.=55.000 mm
-@in the d i h c e betww toI-nq,d aljomnce~
-., .J15#".:Why do m~rm&n~pro&k produce $at& d t b such a wide range af tolerances?
- 33. Thejoumal and M n gwembljr has abasii sizeof 206m.For thebearing :
ED.; (Here, L.D.) =0 micron, and Toterance=46 micron.
For thejournal, the values are-820 and 115 micronsrespectively. F a tfiG4- o f j d
and bearing, assumingunilateralsystem of tolerances.Alp, +tpminingtheallowanceandtypeof
'" ,.' '; ''tit. Which spkms of fithas,knadopted? < .
. .I ' 4 ijl$ I h b: 'J'fl5.? & ' L ~ P , .~fc3
Minimum sizeof the hole = 3 & h '
< . I 9 <.
W.Maximum@6f t.k shaQ = 50 mm. .--a gihwrtC .
Minimum sizesftheshaft = 50-0.01 1=49.989 mm +i@@q01 gfribo23 "
Maximum sizeof thehole = 50-0 . q =49.974 w f:, --t (,, t,,,,., ,Minimum sizeofthe hole = 50-0.065 =49.935 mm -
MaximumClearance = Maximum size of hole -Minimm sizeofBe'shaft
- -49.9% -W.989--0.01!? nun
M i n i m u m C l m = M4nbum&e of hole -W m u m sizaof the shaa
= 49.935 -50~000=-0.065 llun
Sincebdfithemaximumandmhiiarrpncl- ann w v e , thefit obtrdnedwill be "ineaEemnce
fit" :,: ,%,k3 36(! ;,,lii . 1 t P % 4 38,
36. Determinethe typeof fit :
.
Hale size : + 0'05 mm ; S ~ B:20
20 -0.05
+ 0.05 -9-3' A .,-0.05 ['IS ' :
@huv sizeofthehole = 20.05 - , * , ; , , *MinimwsizeoPtheQole = 19.95pun
. ~ c > ; q > < 13; Id,(* ..J& ,-rJy ?I{: : ..j&#(bL:ai
i X
MaximumQzeof the shaft = 20.05 mm r
* . ' k ; ( ~ ~ ~ ' d ~Tm~6 = 19.* - :31 qd t t l d ~ ~ ~ki4d.;.:>/~*:*>I: fii .#ficil ,:h
i
~ ~ ~ d i a t o f k d e - M i n & n r m ~ . o f t b M * Z @ . @ 5 -1935~0,101~~n
sw efReld- wUIm~ a f ~=19.96-~.055-0.10m
positive9ndthe Minimum Clearanceis negative, the titobtainJ
1
, . .I
-* - - - f

11. Gauges and Gauge Design 407
(iii) Combined Bore / Face Gauge: The position and parallelism of a bore in relation to a
datum face can be checked by means of a combined bore/face gauge, Fig. 10.38. The pin which
locates in the bore is in effect, 'Go' plug gauge, and the steps ground on the other pin are the
'Go' and 'Not Go' limits for the datum face to hole axis dimension. The length of the plug gauge
needs to be sufficient to enable the length of the 'Go' step on the pin to check for parallelism.
The tolerance on the hole must be less than the tolerance on the dimension to the face for the
gauge to operate satisfactorily.
Fig, 10.38. A combined bore/face gauge.
10.11. SOLVED EXAMPLES
Example I. A 25 mm H8--j7 fit is to be checked. The limits oj size Jar H8 hole are: High
limit 25.033 mm, low limit 25.000 mm. The limits of size Jar 17 shafts are: High limit 24.980
mm. low limit 24.959 mm. Taking gauge maker's tolerance to be 10% oj the work tolerance.
design plug gauge and gap gauge to check the fit.
Solution. Tolerance for hole = H.L - L.L.
= 25.033 - 25.000 = 0.033 mm
.. Gauge makers tolerance for plug gauge = 0.1 x 0.033 mm- = 0.0033 mm
= 0.003 mm (rationalised)
Gauge makers tolerance for gap gauge = 0.0021 mm = 0.002 mm (rationalised)
As the work tolerances are less than 0.09 mm, wear allowance may not be provided.
(i) Plug Gauge
Basic size of 'Go' plug gauge = L.L. of the hole (MMC) = 25.000 mm
.'. In unilateral system,
+ 0.003
Dimensions of 'Go' plug gauge = 25.00 mm
- 0.000
That is,
High limit of 'Go' plug gauge = 25.000 + 0.003
= 25.003 mm
Low limit of 'Go' plug gauge = 25.000 mm
Now,
Basic size of 'Not Go' plug gauge = 25.033 mm
+ 0.000
•• Dimensions of 'Not Go' plug gauge = 25.033 mm
- 0.003
(Fig. 10.40 shows a sketch of combined 'Go' and 'Not Go' plug gauge.)

12. -
..,c* . ._A.
. it.
, ', ,,,.: , -/ 4 . . 4 =
,-;,I * - 4 ,
ft8 -&iu
.. I 8 , , I . 0 0 Bt3 I. 1 ^- r . - , . t + , '
, *t .
+ I -.-.---.-.-.-. -....-.-.-.-.-.-.-.-.-.-. -.-.-.- .-.-.-.-.-.-....--.-.-.-.-.--.- iti
, @'
'!!
+$,h - l
r I .
. ; , $.'?.%? 'ah
Fig. 10.40. Plug Gauge (combined type)
(ii)lCap Gauge
'Go' side = H.L.of shaft (MMC)
= 24.980 mm.
:. Dimensions of 'Go" gap gauge = 24.980 rnq
'Not Go' side = L.L. of shaft = 2 4 . 9 5 6 '
+ 0.002
:. Dimensions of 'Not Go' gap gauge = 24.959 mm
b*i@-w
(Fig. 10.41 shows a sketch of combinid 'Go' and 'Not Go' gap gauge)
/
, i . , c ; * ' <
P '
Fig. 10.41. Gap Gauge (combined type).
~ramptd2.!b&h of'732b!@'km'&thew are 'hiBe ~6kkkctby the Ire& of a Go, Not
Go snap gauges. Design the gauge, sketch it and show its Go size'& Not go Y& dimensions.
Assume normal wear allowance and m e maker's tolerance.
Soluti~n. Highlimitofshaft=7d.@2l- -,---.2/111r1 ' . I 1.,11r2t?r~~11~i
Low limit of &Rri 74.98 mm
Work tolerance = 75.02 - 74.98 -0.04 mm
:. Gauge makers tolerance (lO?!Y = 0.& mm - . ,
Wear tolerance = 0.001 mm
'Go side' of snap gauge = H.L.of shaft; (MMC) = 75.02 mm /
'Not Go' side of snap gauge = 74.98 mm
~ i ~ r~ r l
Wear allowanceis to be applied firstto 'GO' ride, before gauge maker's tolarnw is applied.
(Refer to Fig. 10.8).
! ' 1 G I i!
'Go' side of snap gauge afterconsidering the wear allowance
= 75-02-0.0028 . = 75,018 mm ,,,,:,(;k.,jl ,,i l
:. Dirnensi~mof smp gauge are given RS :

13. Unilateral System
+ 0.000
'Go' 75.018 mm
- 0.004
Bilateral System
+ 0.002
'go' 75.018 mm
- 0.002
,,.... ..
'Not Go' 74.98 mm
rnrn FC1 I) 3rl,rk ,,, r., 1-i ltci x ~ r m ~ ! ~ ~ --6.002
k z '-"
Example 3. Find the 'Go' and 'Not Go' gauge dim&%; bTuafihg gauge m%g'8ilateral
and Unilateral System and including wear.allowmce&r gauging 75 =t-(Lp(.my.~ t ~ r J 1 0 1 e s .
soljtion. High limit of hole = 75.05I t * ,m. ' 91o~i10 t I l ! i i i ~ 1o.3
Low limit of hole = 74.95 mm spue;> 0;)
work tolerance = 75.05 - 74.95 = 0.1 mm si~x43ttt.,bwst. Oh (11
Gauge maker's tolerance = 0.01 mm
.s,,r~,@i m#.~1b,&-&~rd!ao5mi,, UOI US br1oq4~1m~Ilia 3 ~ ~ 2na
'm'side of plug ga~gBAf..c.'ofBole 2~a.W'"'
*I2 -
- -
= ~74.9$'hid
6lie&r*$ .of w & $ ~ ~ i + m ~ y ~1ms%Sku qnraif
= 7$.%Z0+ 0.005 (Fig. 10.8)
= 75.955 &&
I
QOl] t.3 -
Dimension of plug gauge are given as .
Unilateral System
woiztrsr&h tn4n 0%fiil
+ 0.010 n o i ? + bmm 08 161 'sgusl) oi)' 'hp ~ - 0 0 0
'Go' 74.955 mm mm.08 = %ot Go' 75.05 mm
- 0.000 w g mi-& lot 'quai3 03'fa s~ - 0-010
Bilateral System traS.0 +
tPs2&0-?Q5 :I m ~ nW.B8 + 0.005
'Go' 74.9H'mm W)O.O - 'Not go' 75.05 mm
- 0.005 FFr fff dq mi -cr 01
-0.005J , , . * .
Example 4. The rectangular hole shown in Fig. 10.43 @ a be checked
The "mi* of size for the 'wt?& "+ 0.04
80 mm
- 0.00
4.04
Design the suitabb gauges ?used on Taylor's * -0.00 m
principle. W m
Solution. Accordingto ~aylor'sprincipfe, there will
a -- Fig.,10.42 -.
,- -- -+be one 'Go Gauge' of full form@a@length e q dtothe ;'-' - . "J:,
length of the hole. Therewill be'two 'Not Go Gauges;' of p&Z$rm to check the width andbreaMh
of the hole. , *vT.=.* &-5
!I

14. A Textbook of f k l ~ t i o nEnginewing
High limit of width of hole = 60.04 mm
Low limit of width of hole = 60.00 mm
High limit of breadth of hole = 80.05 mm
Low limit of breadth of hole = 80.00 mm
Tolerance on width of hole = 0.04 mm
Toleran~eon breadth of hole = 0.05 mm
,,.,,! Gauge marker's t o h n c e (10%) : . , I 1: L L . ~ . t ~~!qt::?c P
"'!' Fot brdtb of hole o.m*.-,4 %!,.hL . ,,< , . I - 1 ) , ,, ., t , 1 , ~ ' 6
sk+d?Q Yk,;~:b f ! > l ( i - iitt! I;<
For width of hole = 0.00i mm
d3LJ fo ftnlll r I.;
Go Gauge ~r - ijr),?T c qnsdoi AX J.,
(i) 60 mm dimension :
Go gauge will correspond to low limit of )vdthof h ~ i e(MYC),i.e., 60 mm.
:. Basic size of 'Go Gauge' f i 60 ,wpimnsion
y 'to h i e c;
= 60.00 rpm
Using unilateral system, the limits of siye for 'Go Gaugq' fa,@:mp dhension are :
1a.t:' . + 0.004
, 60.00 mm
, - 0.000 . -
(io 80 mm dikemion : ' 4 9 .$+ ;
Basier* of 'Go Gauge' for 80 mm dimension
= 80.00 mm
:. L i d b of size of 'Go Gauge' for 80 mm dimension are :
+ 0.005
80.00 mm (unilateral system)
-0.000
The 'bbGauge' is- shown in Fig. 10.43.
:..ilnr-..i
Fig. 10.43.
i
I E

15. Not Go Gauges (Refer Fig. 10.7)
'4 7, '
The 'Not Go Gaugesywill correspond to the mhimum met?! cuwjitim of hole i.e. high
limits of dimensions.
. ,,:. Basic she of 'NoT Go Gauge' for 50 mm dimwshn nri 1~3.: I ?:;me :~rclr? c .-r
= 60.04 rn : .-w :li . I Q I Z ~ : .arg:rFOC
:. Limits of size of 'Not Go Gauge' for 60 mm dimension are zfia.fi 'bdq
- ..- . I-. L.,. _ _ A . -
Limits of size of 'No Go 6'+& P i 80 Y*dimension.
. are :
:6 ?-1. ., .. . : - . 4 . ' ir: ...i.c..,- -*. . .--... ..-.....2 lo 'i~fq~f;:+i-ilj~ l ?
+awl:& + (pi-
1. What is a gauge?
Oi*" + 81,807 &tp pi3 4
2. How dots a gauge differ from a rheasuring instrum&
-4. a! kl 3. --**! Sre - . 4 s o l & ~ ,baPu i w d n.sl;/. IGI-I.,'1 1 ' .9joll(
,.,*. 6 , I l . .,..,v
4. What is the di&tonce between standard gauge (non-limit page) and a limit e?airid31 ' " I , - .,.- ) I 2 ,.It ' I , T1 % ..
ltadZ .+Q* :H : # Q P , P ~ i ~ p w i i ~ n&iie ~d -r 0w0. *'
94: ira 6. What h mu@ ~ d W stdimnce?5Mdw,is ibopaedia the BdsigB of gauges?
7. Give the advantages and disadvantages of unillttellll and bilateral sym of -fig.
'[I,. 8. ww is wcar'dl~w-'? EBW ib itragtrlied in the4dcsignofgitiges? z,
, ,
9. State and explain Taylor's principlesf &it gauging.
, , ,.I& Discuss the.vllrim$nar*erbksu&?d for gaw imw8bEture.
11. What are the advantages of limit gauges'? ..- - , . ,
~ U P P42, we kbt limitations of Ilma gauges'? .c I - ' t! -::.ri . . , , . a ,
13. How the gkgds skula bc c a d for btfore and bRh u3eq' 2 : " '
14. What is M W I I , ,I5 ? , + , , r , a d
15. Determine the specification &$ the 'gdlv a d "n6r 'ga""eiids'of a set of manufacturing and
inspection plug gauges to be used in checking a hole with s ' t c t I l C : l l q n b l 3 ~.-* tc'
+ 0.075 dsloit rs.irirrraa L-J!
diameter specification of 25 mm Ai:t* hh:x.,{. a

16. 002 ' << > . $
16. A (haft I00 + mm diameter is to be checked using a 'go-not go' sna! F g e . Give its
ayr!! 3.
dimensions. Anow for wear &d b g e mak&rlstbikrance. r %Yaitaij C Y ~ $ r SP!
? , i < l i r¶-f.,f, -:, / , I l l , :
17. A limit gauge is required a @he&the hPle 50 :::ma (50 H i , The depth of hole is
200 mm. Design the gauge and sketch it with dimensions.
18. Discuss briefly various aspea ttkr W i n g the td-B on limit gauges. 3.1, ,, :.+:
19. A hole and shaft system has the following dimensions :
. -60 mm HWc8 -pjrrrl'J r,q 1;d lo? '3;uhl: 0 3 ) 13d-5r; >!.I. ?:.'F,.~
The standard tolerance is given by ~.:0.()-i = 0.45 (D)lr3+ O.@I D I
Where D = Diameter of geometric mean of stups. mm
i = stanw tolerance, micron
The multiplier for grade 8 is 25.'The fund&ental, d,yiation for shaft c for D > 40,is given by
- (95 + 0.8 D)
The diameter range lies between 50 to 80 mm. %ketchthe fit and show on it the actual
dimensions of hole and shaft. b e the class aPfit. Also,design the suitable gauges to check
the hole and the shaft (AMIE 1974 S)
[Ans. - - le :LL,= 60.000 mm, H.L.= 59.046 mm
'
Shaft : H.L9-59.854mm, LL.= 59.808 rnm, Clearance fit..x.
.: -- --- - - ' I
. .'-" - -
-- (j";io5;
plug ~aude,GO'& ; 60*0.005~mm
, bd '4
0.000Not Go side : 60.04~-
0.005 mm CLIl *.
Snap Gauge, Go side 59.854f k g
Note. Unilateral system has been used. Wear allowance has been neglected; work taYqace being less
than $09 mm;,,, $ + I ! & -,. '!
20. The minimum size of a hole is 25.00 mm. Its maximum size is 25.002 mm. When matching
shaft ,8fwrldldis &the fbnMrZal Beviation is found to be - 0.02 mm. Shaft
toler&W@ O H 3 m. Ddga &es for hde snd.'shft. TaPce the usual valiles.of gauge
qabfs,tokwe and wear Jlowyrcc.
21. Design 'GO' aryl 'W.GQ;,awls,&LpIqg g ~ ~ t g ~ ito wqwm%hd@pf sb28.4XW*%O 14 mm
adopting (a)Unilateral system (b)' Bilateral system. 8 , . l > t' >
22. A bore of :;::! mm dia x 43 ntt~~long isto be &&A. Iksig@'dilZwd~~&IlImknsior a plug
gauge for this, based on Taylor's principle design. vbf, r-YE ::ti'/.' .!'
23. A square peg having limits of 25.00 nun and 24.97 mm is to ke dPw:W. Design a gauge
(gauges) for checking this,W op Tqyhr's principle ~pfwge dyign. ,, .,,,- .
24. Discuss the principle of Taylor's for the design of gauges for checking i,, , ad
X i2r[IS , ( ~ j . ~ t ~ . ~ ~ d4016 with a c~rilylrical:Not ,Go' g+ge. 1 ~ 3 i t t ~ ~ r '~ o , . ~ ~ ~ ~ . , ~ ~F t
(b) of rc&ngular hole rw xuri g y&;m&, or ~ a z usa 01 b @ ~ : : g21~!r?ourr,x:ln.
(c) circular holes- ?irJuL
(d) circular shafts. rrtnt ?r ?c. itot~:i3r!1?3~jd
(e) Non-circular holes and shafts. 'OC '-
vwi-"

17. Gauges and Gauge Design
25. Sketch and discuss various types of Plug gauges
,.a.Sketch and discuss various types of snap gauges.
$7. Discuss the procedure df manufacturing Limit plug gauges.
28. Discuss the procedure of manufacturing Limit snap gauges.
29. What role do gauges play in the mass production system?
,4
f .
I*'
8
- 31. m a r e Snap geugesw?
--..----- . - -
32. Sketch and discuss the ,mof following gauges :
(a) Length gauges Y ~ C ) ~ - ~ ~ V U ~ . I H ~8 i .I!
- - -.
,"!I I I t;t
(c) Receiver gauge Ern s f l .b.~nrardo 1:~:. :$I.J tm-07 issrr~w~vs$f33!31 I-. if'^
rsq r; 10 *.I . ''f yl~li. tvlwrr -336 rt-,, ,,l)i~b' - rnt 9 ~ t
c (4F'@ pin w % n p-scr' ,ttoi51 to l?lZm3 l - & l ! { z i ~ 4 3 93L.13Jr .,
.i 33. What are sorcw gaugeax?,How .am.aEbdy dad to umtrok.the oompk dirncnshw d $ak&ds?
% f i r r . ,1 f
-. hi To ml.1 F r: . I J - ~ > :
3' ~ 5 6 k ' h h * m gJ . ~ r j r t r ~ 1 ofptsq/&bjO d%o. , .,, ,,,a luaa;l:
, . A w o r ~ . ~ ~ & i t a ~ ~ ~ ~ k b . ~ ~ - f r ~ . r i ~ ~
)-s;~ 'I d*. jkpi kge914 G&uhte.'L' &em 49 1°-L58j,Max. did, = 23.42 nnn, min.
,I,- jAlfigqdia 7 1 3 . 4 Q t ~ .7 . , [ b hb.* (D - dY2 an @In; 0 . M ~m]
1 ~ 5 1 37. w b on wearfar gar$eslakrgsni v.I .z!fizlwsi t~J! a: bshrc
yZ-,-- ._ . . - I 1 . I .-.
"b313: S.I. fi='va&us t n h k I+,I+'- mWI&~ n .10.30), Special W=
mamiets lir q$ktimm W'&kr"~iL~its rik very rapidly ctuc to
wear (for ex~mple,a small diameter screw plug gauge used on a cast iron
*at), t), a :-
.-.(a) h s . (B& w& resistawe than c-stee~). .. .,
&4abpowder dbldhqs ggnrd with d i d grit abrasive, though very
agrasllp.iqmvkl wcar 'life. .--*3I6tr&- -
38. W r b the -UWdvantagesof limit Gauging.
Sol. Two limitatiorrs_oiLliaitG-Aw ,ksndkrseed & Art 10.8. The otha drawbacks
.. .
ntr;tr,;, trt nr-M * nomi c ? f a e i ~ p - ,- I , :. . . .,
->! J B V ~ C (a) .Suitebbfbf Maw~pmitiQI of btjc Theit c a ~myfmbe IB &an
a : 11' a5 ~ i ~ ~ ~ o f d,.> l b n 6 2 : , .: tl3OSg 12.)TT&Yl!f?l Sf*
~ f !(.. )I
(byfie>* of limit gaw h sizes
I F 1 1. :con&&&, s s&H7asTWght~ffb$& i w r f l m e n w g '
16 1 ~ 1 ,r
.-.,,,,,,,,fi,&'kiih MY iimitititWs Ep~bbS
~ ~ * + ~ ~ ' ~ , ~ e ' ' @ ~ , d ~ ~ w ~ d u ~ g b ~i ~ , . , ~ ~ j , , ; a : x : R ~ l , p Jtur: ;!j !-33 a
(d) Particular sources bf error in themampnukt an ~~Irrvs(rk$ Re&wms in machine
>(j <i c Y ~ 3 ,
tQol -4 q d better P t M (. , ' .& due, Q ekqmjx) haw a h ~ ~ dthe
, d ~ s b a y . b , ~ l ; # . mw T + whmg -.- i n e n ulr.I L H L C , !I
I . .?:qr ~~$i't","~,~at$$sTCF~LT;rf lis82
c h d g plain bores, both intemal,andex& somn thrrrds, s#hes and serrations
, r t